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Sarnased probleemid veebiotsingust

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\frac{\frac{\sqrt{5}}{2\sqrt{2}+\sqrt{5}}}{\sqrt{8}-\sqrt{5}}
Tegurda 8=2^{2}\times 2. Kirjutage \sqrt{2^{2}\times 2} toote juured, kui see ruut \sqrt{2^{2}}\sqrt{2}. Leidke 2^{2} ruutjuur.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{\left(2\sqrt{2}+\sqrt{5}\right)\left(2\sqrt{2}-\sqrt{5}\right)}}{\sqrt{8}-\sqrt{5}}
Ratsionaliseerige korrutades lugeja ja 2\sqrt{2}-\sqrt{5} nimetaja \frac{\sqrt{5}}{2\sqrt{2}+\sqrt{5}} nimetaja.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{\left(2\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}}}{\sqrt{8}-\sqrt{5}}
Mõelge valemile \left(2\sqrt{2}+\sqrt{5}\right)\left(2\sqrt{2}-\sqrt{5}\right). Korrutustehte saab ruutude vaheks teisendada järgmise reegli abil: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{2^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}}}{\sqrt{8}-\sqrt{5}}
Laiendage \left(2\sqrt{2}\right)^{2}.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{4\left(\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}}}{\sqrt{8}-\sqrt{5}}
Arvutage 2 aste 2 ja leidke 4.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{4\times 2-\left(\sqrt{5}\right)^{2}}}{\sqrt{8}-\sqrt{5}}
\sqrt{2} ruut on 2.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{8-\left(\sqrt{5}\right)^{2}}}{\sqrt{8}-\sqrt{5}}
Korrutage 4 ja 2, et leida 8.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{8-5}}{\sqrt{8}-\sqrt{5}}
\sqrt{5} ruut on 5.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{3}}{\sqrt{8}-\sqrt{5}}
Lahutage 5 väärtusest 8, et leida 3.
\frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{3}}{2\sqrt{2}-\sqrt{5}}
Tegurda 8=2^{2}\times 2. Kirjutage \sqrt{2^{2}\times 2} toote juured, kui see ruut \sqrt{2^{2}}\sqrt{2}. Leidke 2^{2} ruutjuur.
\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{3\left(2\sqrt{2}-\sqrt{5}\right)}
Avaldage \frac{\frac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{3}}{2\sqrt{2}-\sqrt{5}} ühe murdarvuna.
\frac{\sqrt{5}}{3}
Taandage -\sqrt{5}+2\sqrt{2} nii lugejas kui ka nimetajas.